# HW4 - 14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK...

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14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK 04 Question 5.3: ( a) Compare interstitial and vacancy atomic mechanisms for diffusion. (b) Cite two reasons why interstitial diffusion is normally more rapid than vacancy diffusion. Solution of 5.3: (a) With vacancy diffusion, atomic motion is from one lattice site to an adjacent vacancy. Self-diffusion and the diffusion of substitutional impurities proceed via this mechanism. On the other hand, atomic motion is from interstitial site to adjacent interstitial site for the interstitial diffusion mechanism. (b) Interstitial diffusion is normally more rapid than vacancy diffusion because: (1) interstitial atoms, being smaller, are more mobile; and (2) the probability of an empty adjacent interstitial site is greater than for a vacancy adjacent to a host (or substitutional impurity) atom. Question 5.4: Briefly explain the concept of steady state as it applies to diffusion. Solution of 5.4: Steady-state diffusion is the situation wherein the rate of diffusion into a given system is just equal to the rate of diffusion out, such that there is no net accumulation or depletion of diffusing species-i.e., the diffusion flux is independent of time. Question 5.6: The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3. Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having an area of 0.20 m 2 at 500 C. Assume a diffusion coefficient of 1.0 10 -8 m 2 /s, that the concentrations at the high- and low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steady- state conditions have been attained. Solution of 5.6: This problem calls for the mass of hydrogen, per hour, that diffuses through a Pd sheet. It first becomes necessary to employ both Equations 5.1a and 5.3. Combining these expressions and solving for the mass yields M = JAt DAt C x (1.0 10 -8 m 2 /s )(0.20 m 2 )(3600 s/h) 0.6 2.4 kg /m 3 5 10 3 m = 2.6 10 -3 kg/h

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Question 5.8: A sheet of BCC iron 1 mm thick was exposed to a carburizing gas atmosphere on one side and a decarburizing atmosphere on the other side at 725 C. After having reached steady state, the iron was quickly cooled to room temperature. The carbon concentrations at the two surfaces of the sheet were determined to be 0.012 and 0.0075 wt%. Compute the diffusion coefficient if the diffusion flux is 1.4 10 -8 kg/m 2 -s. Hint: Use Equation 4.9 to convert the concentrations from weight percent to kilograms of carbon per cubic meter of iron. Solution of 5.8: Let us first convert the carbon concentrations from weight percent to kilograms carbon per meter cubed using Equation 4.9a. For 0.012 wt% C C C " = C C C C C C Fe Fe 10 3 0.012 0.012 2.25 g/cm 3 99.988 7.87 g/cm 3 10 3 0.944 kg C/m 3 Similarly, for 0.0075 wt% C C C " 0.0075 0.0075 2.25 g/cm 3 99.9925 7.87 g/cm 3 10 3 = 0.590 kg C/m 3 Now, using a rearranged form of Equation 5.3 D J x A x B C A C B ( 1.40 10 -8 kg/m 2 -s ) 10 3 m 0.944 kg/m 3 0.590 kg/m 3
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HW4 - 14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK...

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